Vectors Angle Cosine 1
Let $ \mathbf{u} $ and $ \mathbf{v} $ be vectors such that $ \|\mathbf{u}\| = \|\mathbf{v}\| = 2 $ and $ \mathbf{u} \cdot \mathbf{v} = -1 $. If $ \theta $ is the angle between the vectors $ \mathbf{u} + \mathbf{v} $ and $ 2 \mathbf{u} - \mathbf{v}, $ then find $ \cos \theta $.
- 1
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- +
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- -
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- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
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- $[$
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- $\cap$
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- $\infty$
Solution
Note that
\begin{align*}
(\mathbf{u} + \mathbf{v}) \cdot (2 \mathbf{u} - \mathbf{v}) &= 2 \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{v} \\
&= 2 \cdot 2^2 + (-1) - 2^2 \\
&= 3.\end{align*}Also,
\begin{align*}
\|\mathbf{u} + \mathbf{v}\| &= \sqrt{(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})} \\
&= \sqrt{\mathbf{u} \cdot \mathbf{u} + 2 \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v}} \\
&= \sqrt{2^2 + 2(-1) + 2^2} \\
&= \sqrt{6},
\end{align*}and
\begin{align*}
\|2 \mathbf{u} - \mathbf{v}\| &= \sqrt{(2 \mathbf{u} - \mathbf{v}) \cdot (2 \mathbf{u} - \mathbf{v})} \\
&= \sqrt{4 \mathbf{u} \cdot \mathbf{u} - 4 \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v}} \\
&= \sqrt{4 \cdot 2^2 - 4(-1) + 2^2} \\
&= \sqrt{24} = 2 \sqrt{6}.\end{align*}Hence,
\[\cos \theta = \frac{(\mathbf{u} + \mathbf{v}) \cdot (2 \mathbf{u} - \mathbf{v})}{\|\mathbf{u} + \mathbf{v}\| \|2 \mathbf{u} - \mathbf{v}\|} = \frac{3}{\sqrt{6} \cdot 2 \sqrt{6}} = \boxed{\frac{1}{4}}.\]
Freeflows